Solutions - Exercise 1 (C) | Number System | Class 6 Mathematics | Selina Concise Mathematics - Middle School.
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📂 Chapter 1 - Exercise 1 (C).
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# Units of Length, Mass and Capacity :
Unit means single thing e.g. a litre, a day, a minute, a kilometre, etc.
The standard unit of length, weight (mass) and capacity (volume) are metre (m), gram (g) and litre (L) respectively.
| Units of | Length | Units of | Mass | Units of | Capacity | ||
|---|---|---|---|---|---|---|---|
| Millimetre (mm) | 1/1000 metre | Milligram (mg) | 1/1000 gram | Millilitre (mL) | 1/1000 litre | ||
| Centimetre (cm) | 1/100 metre | Centigram (cg) | 1/100 gram | Centilitre (cl) | 1/100 litre | ||
| Decimetre (dm) | 1/10 metre | Decigram (dg) | 1/10 gram | Decilitre (dl) | 1/10 litre | ||
| A metre (m) | Standard unit | A gram (g) | Standard unit | A Litre (L) | Standard unit | ||
| Decametre (dam) | 10 metre | Decagram (dag) | 10 gram | Decalitre (dal) | 10 litre | ||
| Hectometre (hm) | 100 metre | Hectogram (hg) | 100 gram | Hectolitre (hl) | 100 litre | ||
| Kilometre (km) | 1000 metre | kilogram (kg) | 1000 gram | Kilolitre (kl) | 1000 litre |
📌 Note : Numbers which do not refer to any particular unit such as 8, 12, 15, 23, etc. are called abstract numbers. whereas, the numbers which refers to a particular unit such as 8L, 12m, 15 days, 23 hours, etc. are called concrete numbers.
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-- EXERCISE 1 (C) --
Q1. Population of a city was 3,54,976 in the year 2014. In the year 2015, it increased by 68,438. What was the population of the city at the end of the year 2015 ?
Solution :
Population of city in the year 2014 = 3,54,976
Population of city increased by in the year 2015 = 68,438
Population of city at the end of the year 2015 = 3,54,976 + 68, 438 = 4,23,414.
Ans. Population of city at the end of the year 2015 is 4,23,414.
Q2. A = 7,43,000 and B = 8,00,100. Which is greater A or B ? And by how much ?
Solution :
The first number is 7,43,000 i.e. 7,00,000 + 40,000 + 3000
The second number is 8,00,100 i.e. 8,00,000 + 100
Here,
The second number 8,00,100 has high place value i.e. 8,00,000
Therefore, 8,00,100 is the greater number.
Difference between the two numbers = 8,00,100 - 7,43,000 = 57,100.
Ans. A is greater by 57,100.
Q3. A notebook has 56 pages. How many pages will 5326 such notebooks have ?
Solution :
Number of pages in one notebook = 56
Number of Notebooks = 5326
Total number of pages in 5326 notebooks = 5326 x 56 = 2,98,256.
Ans. 2,98,256.
Q4. A person has 75,000 sheets of paper. If each sheet makes 8 pages of a notebook, how many notebooks of 200 pages can be made using the above sheets ?
Solution :
Number of sheets of paper a person has = 75,000
Number of pages each sheet can make = 8
Total number of pages obtained = 8 x 75,000 = 6,00,000
Number of pages in each notebook = 200
Total number of notebooks made = 6,00,000 ÷ 200 = 3,000
Ans. Total number of notebooks made out of 6,00,000 pages are 3,000.
Q5. Add 1,76,209 ; 4,50,923 and 44,83,947.
Solution :
Sum of 1,76,209 + 4,50,923 + 44,83,947
1,76,2094,50,923+ 44,83,94751,11,079
Ans. 51,11,079
Q6. A cricket player has scored 7,849 runs so far in test matches. He wishes to complete 10,000 runs; how many more runs does he need ?
Solution :
Runs scored by a cricket player = 7,849
Runs wishes to complete by the cricket player = 10,000
Runs need to be scored by the cricket player = 10,000 - 7849 = 2151
Ans. Runs need to be scored by the cricket player are 2151.
Q7. In an election two candidates A and B are the only contestants. If candidate A scored 9,32,567 votes and candidate B scored 9,00,235 votes, by how much margin did A win or loose the election ?
Solution :
Votes scored by candidate A = 9,32,567
Votes scored by candidate B = 9,00,235
Difference = 9,32,567 - 9,00,235 = 32,332
Ans. A wins by 32,332 votes.
Q8. Find the difference between the largest and the smallest five digit numbers that can be formed using the digits 5, 1, 6, 3 and 2 without repeating any digit.
Solution :
Smallest number = 12356
Greatest number = 65321
Difference = 65321 - 12356 = 52,965
Ans. 52,965
Q9. A machine manufactures 5,782 screws every day. If we round off the number of screws produced in a day to the nearest thousand, approximately how many screws will the machine manufacture in the month of April ?
Solution :
Number of screws manufactured in a day = 5782
Rounding off to the nearest thousand = 6000
Number of days in the month of April = 30
Number of screws manufactured in the month of April = 6000 x 30 = 1,80,000
Ans. Number of screws manufactured in the month of April are 1,80,000
Q10. A man has ₹ 1,57,184 with him. He placed an order for purchasing 80 articles at ₹ 125 each. How much money will remain with him after the purchase ?
Solution :
Amount of money a man has = ₹1,57,184
Cost of one article purchase by the man = ₹125
Number of articles purchase by the man = 80
Money spent on purchase of articles = 80 x 125 = 10,000
Money remain with the man after purchase = 1,57,184 - 10,000 = ₹1,47,184
Ans. Money remain with the man after purchase of articles is ₹1,47,184
Q11. A student multiplied 8,035 by 87 instead of multiplying by 78. By how much was his answer greater than or less than the correct answer ?
Solution :
Number to be multiplied = 8,035
Number by which it multiplied = 87
Number by which it should be multiplied = 78
Difference between the two numbers = 87-78 = 9
Difference between the answer = 9 x 8,035 = 72,315
Ans. The answer is greater by 72,315.
Alternate Method :
Number to be multiplied = 8,035
Number by which it multiplied = 87
Multiplication = 8,035 x 87 = 6,99,045
Number by which it should be multiplied = 78
Multiplication = 8,035 x 78 = 6,26,730
Difference between the answer = 6,99,045 - 6,26,730 = 72,315
Ans. The answer is greater by 72,315.
Q12. Mohani has 30 m cloth and she wants to make some shirts for her son. If each shirt requires 2 m 30 cm cloth, how many shirts, in all, can be made and how much length of the cloth will be left ?
Solution :
Length of cloth Mohani has = 30m = 3000cm
Length of cloth required to make a shirt = 2m 30cm = 230cm
Number of shirts Mohani can made = 3000 ÷ 230 = 13
Cloth remained = 10 cm
Ans. Mohani can make 13 shirts and will left with 10 cm of cloth.
Q13. The weight of a box is 4 kg 859 g. What will be the approximate weight of 150 such boxes if we round off the weight of each box to 5 kg ?
Solution :
Weight of one box = 4 kg 859g
Rounding off the weight of one box = 5 kg
Number of boxes = 150
Total weight of all boxes = 150 x 5 = 750kg
Ans. The approximate weight of 150 boxes is 750kg.
Q14. The distance between two places A and B is 3 km 760 m. A boy travels from A to B and then B to A every day. How much distance does he travel in 8 days ?
Solution :
Distance between the two points = 3km 760m = 3760m
Distance travelled in one day = 3760 x 2 = 7520m
Number of days he travelled = 8
Total distance travelled by the boy = 7520 x 8 = 60,160m = 60km 160m
Ans. Distance travelled by the boy in days is 60km 160m.
Q15. An oil-tin contains 6 litre and 60 ml oil. How many identical bottles can this oil fill, if the capacity of each bottle is 30 ml ?
Solution :
Quantity of oil a tin contains = 6L 60ml = 6060ml
Capacity of each bottle = 30ml
Total number of bottle that can be filled = 6060 ÷ 30 = 202
Ans. Number of bottles that can be filled are 202.
Q16. The sale receipt of a company in a certain year was ₹ 83,73,540. In the following year, it decreased by ₹ 7,84,670.
(i) What was the sale receipt of the company during the following year ?
Solution :
Sale receipt of a company in first year = ₹83,73,540
Decrease in sale receipt of the company in following year = ₹7,84,670
Sale receipt of the company in the following year = 83,73,540 - 7,84,670 = ₹75,88,870
Ans. Sale receipt of the company in the following year is ₹75,88,870
(ii) What was the total sale receipt of the company during these two years ?
Solution :
Sale receipt of a company in first year = ₹83,73,540
Sale receipt of the company in the following year = ₹75,88,870
Total sale receipt of the company during these two years = 83,73,540 + 75,88,870 = 1,59,62,410.
Ans. Total sale receipt of the company during these two years is 1,59,62,410.
Q17. A number exceeds 8,59,470 by 3,00,999. What is the number ?
Solution :
First number = 8,59,470
Difference in two numbers = 3,00,999
Second number = 8,59,470 + 3,00,999 = 11,60,469
Ans. The second number is 11,60,469.
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